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JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Sequences and Series

\frac{2 ^{3} - 1 ^{3}}{1 \times 7} + \frac{4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{2 \times 11} + \frac{6 ^{3} - 5 ^{3} + 4 ^{3} - 3 ^{3} + 2 ^{3} - 1 ^{3}}{3 \times 15} + \dots + \frac{3 0 ^{3} - 2 9 ^{3} + 2 8 ^{3} - 2 7 ^{3} + \dots + 2 ^{3} - 1 ^{3}}{15 \times 63}

is equal to

Answer: 120

Solution

The general term is

T_{n} = \frac{2 \sum _{r = 1}^{n} ( 2 r ) ^{3} - \sum _{r = 1}^{2 n} r ^{3}}{n ( 4 n + 3 )}

. Using the formulas for sum of cubes:

\sum_{r = 1}^{n} (2 r)^{3} = 8 (\frac{n ( n + 1 )}{2})^{2} = 2 n^{2} (n + 1)^{2}

and

\sum_{r = 1}^{2 n} r^{3} = n^{2} (2 n + 1)^{2}

. Substituting and simplifying:

T_{n} = \frac{4 n ^{2} ( n + 1 ) ^{2} - n ^{2} ( 2 n + 1 ) ^{2}}{n ( 4 n + 3 )} = n

. Therefore

\sum_{n = 1}^{15} T_{n} = \sum_{n = 1}^{15} n = \frac{15 \times 16}{2} = 120

.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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